- MATH1502 G Group Calculus II Final Exam Corrected Wednesday May 5th 2010 2:50 5:40 No calculators or notes are allowed There are 140 marks on

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Unformatted text preview: MATH1502 - G Group Calculus II Final Exam Corrected Wednesday May 5th, 2010, 2:50 - 5:40 No calculators or notes are allowed. There are 140 marks on the &nal exam, 130=100%. Name_______________________________________ Student Number ______________________________ Group G ___________________________________ TA _________________________ ______________ Question Grade Out of 1 14 2 15 3 24 4 40 5 47 Total 140 1 Question 1 (a) Find the radius of convergence and interval of convergence of the power series 1 X k =2 1 & 1 3 cos k& ( k + 1) 7 = 5 (14 x ) k : (8 marks) Solution We use the root test applied to a k = & & & & & 1 & 1 3 cos k& ( k + 1) 7 = 5 (14 x ) k & & & & & = 1 & 1 3 cos k& ( k + 1) 7 = 5 (14 j x j ) k : (Recall j cos k& j ¡ 1 ). We see that lim k !1 a 1 =k k = lim k !1 1 & 1 3 cos k& ( k + 1) 7 = 5 (14 j x j ) k ! 1 =k = lim k !1 1 & 1 3 cos k& ( k + 1) 7 = 5 ! 1 =k (14 j x j ) = 14 j x j : (2 marks) ( Justi&cation of this, not really required from students: Recall lim k !1 k 1 =k = 1 , so 2 3 1 k 7 = 5 ¡ 1 + 1 k ¢ 7 = 5 ¡ 1 & 1 3 cos k& ( k + 1) 7 = 5 ¡ 4 3 1 k 7 = 5 : and then £ 2 3 ¤ 1 =k 1 ( k 1 =k ) 7 = 5 ¡ 1 + 1 k ¢ 7 = (5 k ) ¡ 1 & 1 3 cos k& ( k + 1) 7 = 5 ! 1 =k ¡ £ 4 3 ¤ 1 =k 1 ( k 1 =k ) 7 = 5 : By the pinching/ sandwich rule, and as lim k !1 ¡ 4 3 ¢ 1 =k = 1 and lim k !1 ¡ 1 + 1 k ¢ 1 =k = 1 ; lim k !1 1 & 1 3 cos k& ( k + 1) 7 = 5 ! 1 =k = 1 : ) 2 By the root test, we have convergence for j x j < 1 14 : And we have divergence for j x j > 1 . So the radius of convergence is 1 14 . (2 marks) Next, we test the endpoints x = & 1 14 . x = ¡ 1 14 Here 1 X k =2 1 ¡ 1 3 cos k& ( k + 1) 7 = 5 (14 x ) k = 1 X k =2 1 ¡ 1 3 cos k& ( k + 1) 7 = 5 ( ¡ 1) k : Now taking absolute values, & & & & & 1 ¡ 1 3 cos k& ( k + 1) 7 = 5 ( ¡ 1) k & & & & & ¢ 4 3 k 7 = 5 and note that the series P 1 k 7 = 5 converges ( p ¡ series with p = 7 5 > 1 ). The comparison test shows that 1 X k =2 1 ¡ 1 3 cos k& ( k + 1) 7 = 5 ( ¡ 1) k converges absolutely, and hence converges. x = 1 14 Here again 1 X k =2 1 ¡ 1 3 cos k& ( k + 1) 7 = 5 (14 x ) k = 1 X k =2 1 ¡ 1 3 cos k& ( k + 1) 7 = 5 and the argument used above shows that the series converges absolutely, and hence converges. So the interval of convergence is ¡ ¡ 1 14 ; 1 14 ¢ : (4 marks) 3 (b) Find the radius of convergence of the power series 1 X k =0 ( k !) 2 (3 k ) 2 k x k : (6 marks) You may assume the limit lim k !1 & 1 + 1 k ¡ k = e: Solution Because of the factorials, we use the ratio test, applied to a k = ¢ ¢ ¢ ¢ ¢ ( k !) 2 (3 k ) 2 k x k ¢ ¢ ¢ ¢ ¢ = ( k !) 2 (3 k ) 2 k j x j k ; with x 6 = 0 . We see that lim k !1 a k +1 a k = lim k !1 (( k +1)!) 2 (3( k +1)) 2 k +2 £ j x j k +1 ¤ ( k !) 2 (3 k ) 2 k j x j k (2 marks) = lim k !1 k 2 k ( k + 1) 2 k +2 3 2 k 3 2 k +2 & ( k + 1)!...
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This note was uploaded on 09/09/2011 for the course MATH 1502 taught by Professor Mcclain during the Fall '07 term at Georgia Institute of Technology.

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- MATH1502 G Group Calculus II Final Exam Corrected Wednesday May 5th 2010 2:50 5:40 No calculators or notes are allowed There are 140 marks on

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